Đáp án:
e) \(\left[ \begin{array}{l}
x = - 1\\
x = - 10
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2\left( {x - \dfrac{1}{2}} \right) = \dfrac{{19}}{{12}}\\
\to x - \dfrac{1}{2} = \dfrac{{19}}{{24}}\\
\to x = \dfrac{{31}}{{24}}\\
c)\left| {\dfrac{2}{3}x - 1} \right| = \dfrac{3}{2}\\
\to \left[ \begin{array}{l}
\dfrac{2}{3}x - 1 = \dfrac{3}{2}\\
\dfrac{2}{3}x - 1 = - \dfrac{3}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{2}{3}x = \dfrac{5}{2}\\
\dfrac{2}{3}x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{15}}{4}\\
x = - \dfrac{3}{4}
\end{array} \right.\\
b)x - 2 = \dfrac{8}{3}\\
\to x = \dfrac{{14}}{3}\\
d){\left( {2x - 1} \right)^2} = 4\\
\to \left| {2x - 1} \right| = 2\\
\to \left[ \begin{array}{l}
2x - 1 = 2\\
2x - 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
e)\left[ \begin{array}{l}
x + 1 = 0\\
\dfrac{1}{2}x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - 10
\end{array} \right.
\end{array}\)
